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Class 9 Mathematics Notes – Chapter 9: Similar Figures

Class: 9th
Subject: Mathematics
Chapter: 09 – Similar Figures
Board: Punjab Board

Exercise 9.1

Q1. Find Whether the Solids Are Similar. All Lengths Are in cm.

Given:

Solid 1: Length = 3 cm, Width = 5 cm, Height = 4 cm
Solid 2: Length = 4.5 cm, Width = 7.5 cm, Height = 6 cm

Step 1: Calculate the ratios of corresponding sides.

Length ratio:

\[ \frac{4.5}{3}=1.5 \]

Width ratio:

\[ \frac{7.5}{5}=1.5 \]

Height ratio:

\[ \frac{6}{4}=1.5 \]

Conclusion: Since all corresponding sides have the same ratio of \(1.5\), the solids are similar.

Q2. Prove That the Triangles Are Similar.

In \(\triangle ABC\), the sides are:

\[ AB=6\text{ cm},\quad BC=15.75\text{ cm},\quad CA=12\text{ cm} \]

In \(\triangle DEF\), the corresponding sides are:

\[ DE=10.5\text{ cm},\quad EF=21\text{ cm},\quad FD=9\text{ cm} \]

Step 1: Find the ratios of corresponding sides.

\[ \frac{DE}{AB}=\frac{10.5}{6}=1.75 \]

\[ \frac{EF}{BC}=\frac{21}{15.75}=\frac43 \]

\[ \frac{FD}{CA}=\frac9{12}=0.75 \]

Note: These ratios are not equal, so the supplied measurements do not establish similarity by the SSS criterion. Check the original textbook dimensions before publishing this question.

Q3. Find the Lengths of \(DF\) and \(EF\).

Given:

\[ \triangle ABC\sim\triangle DEF \]

\[ AB=12\text{ cm},\quad AC=16\text{ cm},\quad BC=20\text{ cm} \]

Perimeter of \(\triangle DEF=30\text{ cm}\) and \(DE=6\text{ cm}\).

Step 1: Find the scale factor.

\[ k=\frac{DE}{AB}=\frac6{12}=\frac12 \]

Step 2: Find \(EF\).

\[ EF=k\times BC \]

\[ EF=\frac12\times20=\boxed{10\text{ cm}} \]

Step 3: Find \(DF\).

\[ DF=k\times AC \]

\[ DF=\frac12\times16=\boxed{8\text{ cm}} \]

Q4. Find the Value of \(x\) in Each of the Following.

(i) Similar Right Triangles

Given:

Smaller triangle: perpendicular = 1.2 cm, base = 3 cm
Larger triangle: perpendicular = \(x\), base = \(3+7.5=10.5\) cm

By similarity of triangles,

\[ \frac{x}{1.2}=\frac{10.5}{3} \]

\[ x=1.2\times3.5 \]

\[ \boxed{x=4.2\text{ cm}} \]

(ii) Parallel Lines Configuration

By the proportional segments theorem,

\[ \frac{x}{6}=\frac38 \]

\[ x=\frac{3\times6}{8} \]

\[ x=\frac{18}{8} \]

\[ \boxed{x=2.25\text{ cm}} \]

(iii) Similar Triangles

By similarity,

\[ \frac{x}{3.6}=\frac{2.5}{2} \]

\[ x=3.6\times1.25 \]

\[ \boxed{x=4.5\text{ cm}} \]

Q5. A Plank Placed Upstairs Is 20 cm Wide and 16 cm Deep. A Box of Height 8 cm and Width \(x\) cm Is Placed Under the Plank. Find \(x\).

The triangles formed by the stairs and the box are similar.

\[ \frac{8}{16}=\frac{20-x}{20} \]

\[ \frac12=\frac{20-x}{20} \]

\[ 20=2(20-x) \]

\[ 20=40-2x \]

\[ 2x=20 \]

\[ \boxed{x=10\text{ cm}} \]

Q6. A Man 1.8 m Tall Casts a Shadow 0.76 m Long. A Telephone Pole Casts a 3 m Shadow. Find the Height of the Pole.

Using the similarity ratio, let the height of the pole be \(h\).

\[ \frac{h}{1.8}=\frac3{0.76} \]

\[ h=\frac{3\times1.8}{0.76} \]

\[ h=\frac{5.4}{0.76} \]

\[ \boxed{h\approx7.11\text{ m}} \]

Q7. Find the Values of \(x\), \(y\), and \(z\) in the Given Right Triangle with an Altitude to the Hypotenuse.

Given: One hypotenuse segment is 6 cm, the altitude is \(y\), and one leg is 10 cm.

Step 1: Find \(x\) using the leg theorem.

\[ 10^2=6(6+x) \]

\[ 100=36+6x \]

\[ 6x=64 \]

\[ \boxed{x=\frac{32}{3}\text{ cm}\approx10.67\text{ cm}} \]

Step 2: Find \(y\) using the geometric mean theorem.

\[ y^2=6x \]

\[ y^2=6\times\frac{32}{3}=64 \]

\[ \boxed{y=8\text{ cm}} \]

Step 3: Find \(z\).

Using the supplied relationship,

\[ z=\sqrt{x^2+y^2} \]

\[ z=\sqrt{\left(\frac{32}{3}\right)^2+8^2} \]

\[ z=\sqrt{\frac{1024}{9}+64} \]

\[ z=\sqrt{\frac{1600}{9}} \]

\[ \boxed{z=\frac{40}{3}\text{ cm}\approx13.33\text{ cm}} \]

Q8. An Isosceles Trapezoid \(ABCD\) Has \(AB\parallel CD\), \(AB=8\) cm, \(CD=4\) cm, and \(AE=3\) cm. Find \(CE\).

Since \(\triangle ABE\sim\triangle CDE\),

\[ \frac{CE}{AE}=\frac{CD}{AB} \]

\[ \frac{CE}{3}=\frac48=\frac12 \]

\[ CE=3\times\frac12 \]

\[ \boxed{CE=1.5\text{ cm}} \]

Q9. A Regular Dodecagon Has Its Side Lengths Decreased by a Factor of \(\frac1{\sqrt2}\). Its Original Perimeter Is 72 cm. Find the Scaled Side Length.

Step 1: Find the original side length.

A regular dodecagon has 12 sides.

\[ \text{Original side length}=\frac{72}{12}=6\text{ cm} \]

Step 2: Apply the scale factor.

\[ \text{Scaled side length}=6\times\frac1{\sqrt2} \]

\[ =\frac6{\sqrt2}=3\sqrt2 \]

\[ \boxed{3\sqrt2\text{ cm}\approx4.24\text{ cm}} \]

Exercise 9.2

Formula:

\[ \frac{A_1}{A_2}=\left(\frac{l_1}{l_2}\right)^2=k^2 \]

Q1. Find the Ratio of Areas if the Ratio of Corresponding Lengths Is:

(i) \(1:3\)

\[ 1^2:3^2=\boxed{1:9} \]

(ii) \(3:4\)

\[ 3^2:4^2=\boxed{9:16} \]

(iii) \(2:7\)

\[ 2^2:7^2=\boxed{4:49} \]

(iv) \(8:9\)

\[ 8^2:9^2=\boxed{64:81} \]

(v) \(6:5\)

\[ 6^2:5^2=\boxed{36:25} \]

Q2. Find the Unknowns in the Following Figures.

(i) Given \(A_1=240\text{ cm}^2\), \(l_1=10\) cm, \(l_2=6\) cm. Find \(A_2\).

\[ \frac{A_1}{A_2}=\left(\frac{l_1}{l_2}\right)^2 \]

\[ \frac{240}{A_2}=\left(\frac{10}{6}\right)^2=\frac{25}{9} \]

\[ A_2=240\times\frac9{25} \]

\[ \boxed{A_2=86.4\text{ cm}^2} \]

(ii) Given \(A_1=60\text{ cm}^2\), \(l_1=15\) cm, \(l_2=20\) cm. Find \(A_2\).

\[ \frac{60}{A_2}=\left(\frac{15}{20}\right)^2=\frac9{16} \]

\[ A_2=60\times\frac{16}{9} \]

\[ \boxed{A_2\approx106.67\text{ cm}^2} \]

(iii) Given \(A_2=18\text{ cm}^2\), \(l_1=3.6\) cm, \(l_2=5.76\) cm. Find \(A_1\).

\[ \frac{A_1}{18}=\left(\frac{3.6}{5.76}\right)^2 \]

\[ =(0.625)^2 \]

\[ A_1=18\times0.390625 \]

\[ \boxed{A_1\approx7.03\text{ cm}^2} \]

(iv) Given \(A_2=96\text{ cm}^2\), \(l_1=15\) cm, \(l_2=12\) cm. Find \(A_1\).

\[ \frac{A_1}{96}=\left(\frac{15}{12}\right)^2=\frac{25}{16} \]

\[ A_1=96\times\frac{25}{16} \]

\[ \boxed{A_1=150\text{ cm}^2} \]

(v) Given \(A_1=14\text{ cm}^2\), \(A_2=63\text{ cm}^2\), \(l_1=3\) cm. Find \(l_2\).

\[ \frac{14}{63}=\left(\frac3{l_2}\right)^2 \]

\[ \frac29=\frac9{l_2^2} \]

\[ 2l_2^2=81 \]

\[ l_2^2=40.5 \]

\[ \boxed{l_2\approx6.36\text{ cm}} \]

Q3. The Area of \(\triangle ABC\) Is 36 cm², \(AB=6\) cm, \(BD=4\) cm, and \(BC\parallel DE\).

Step 1: Find \(AD\).

\[ AD=AB+BD=6+4=10\text{ cm} \]

(a) Find the Area of \(\triangle ADE\).

Since the triangles are similar,

\[ \frac{\text{Area}(\triangle ABC)} {\text{Area}(\triangle ADE)} =\left(\frac{AB}{AD}\right)^2 \]

\[ \frac{36}{\text{Area}(\triangle ADE)} =\left(\frac6{10}\right)^2 =\frac9{25} \]

\[ \text{Area}(\triangle ADE)=36\times\frac{25}{9} \]

\[ \boxed{100\text{ cm}^2} \]

(b) Find the Area of Trapezium \(BCED\).

\[ \text{Area}(BCED)=\text{Area}(\triangle ADE)-\text{Area}(\triangle ABC) \]

\[ =100-36 \]

\[ \boxed{64\text{ cm}^2} \]

Q4. Two Similar Triangles Have a Scale Factor \(k=3\). The Area of \(\triangle ABC\) Is 50 cm². Find the Area of \(\triangle DEF\).

\[ \frac{\text{Area}(\triangle DEF)} {\text{Area}(\triangle ABC)}=k^2 \]

\[ \frac{\text{Area}(\triangle DEF)}{50}=3^2=9 \]

\[ \text{Area}(\triangle DEF)=50\times9 \]

\[ \boxed{450\text{ cm}^2} \]

Q5. Quadrilaterals \(ABCD\sim EFGH\) Have a Scale Factor \(k=\frac14\). The Area of \(ABCD\) Is 64 cm². Find the Area of \(EFGH\).

\[ \text{Area}(EFGH)=k^2\times\text{Area}(ABCD) \]

\[ =\left(\frac14\right)^2\times64 \]

\[ =\frac1{16}\times64 \]

\[ \boxed{4\text{ cm}^2} \]

Q6. The Areas of Two Similar Triangles Are 16 cm² and 25 cm². Find the Ratio of Their Corresponding Sides.

\[ \frac{l_1}{l_2}=\sqrt{\frac{A_1}{A_2}} \]

\[ =\sqrt{\frac{16}{25}}=\frac45 \]

Answer: \(\boxed{4:5}\)

Q7. The Areas of Two Similar Triangles Are 144 cm² and 81 cm². The Base of the Larger Triangle Is 30 cm. Find the Base of the Smaller Triangle.

\[ \frac{\text{Base}_{\text{small}}}{\text{Base}_{\text{large}}} =\sqrt{\frac{81}{144}} =\frac9{12} =\frac34 \]

\[ \text{Base}_{\text{small}}=30\times\frac34 \]

\[ \boxed{22.5\text{ cm}} \]

Q8. The Side of a Regular Heptagon Is 1.7 Times the Side of a Smaller Similar Heptagon. The Area of the Smaller Heptagon Is 100 cm². Find the Area of the Larger Heptagon.

\[ \text{Area}_{\text{large}}=(1.7)^2\times\text{Area}_{\text{small}} \]

\[ =2.89\times100 \]

\[ \boxed{289\text{ cm}^2} \]

Exercise 9.3 Solutions

Q1. The Radii of Two Spheres Are in the Ratio \(3:4\). Find the Ratio of Their Volumes.

For similar solids,

\[ \frac{V_1}{V_2}=\left(\frac{r_1}{r_2}\right)^3 \]

\[ \frac{V_1}{V_2}=\left(\frac34\right)^3=\frac{27}{64} \]

Answer: \(\boxed{27:64}\)

Q2. Two Regular Tetrahedrons Have Volumes in the Ratio \(8:27\). Find the Ratio of Their Sides.

\[ \frac{V_1}{V_2}=\left(\frac{l_1}{l_2}\right)^3 \]

\[ \left(\frac{l_1}{l_2}\right)^3=\frac8{27} \]

Taking cube roots,

\[ \frac{l_1}{l_2}=\sqrt[3]{\frac8{27}}=\frac23 \]

Answer: \(\boxed{2:3}\)

Q3. Two Right Cones Have Volumes in the Ratio \(64:125\). Find the Ratio of Their Heights and Base Areas.

(a) Ratio of Their Heights

\[ \frac{h_1}{h_2}=\sqrt[3]{\frac{V_1}{V_2}} \]

\[ =\sqrt[3]{\frac{64}{125}}=\frac45 \]

Answer: \(\boxed{4:5}\)

(b) Ratio of Their Base Areas

\[ \frac{A_1}{A_2}=\left(\frac{h_1}{h_2}\right)^2 \]

\[ =\left(\frac45\right)^2=\frac{16}{25} \]

Answer: \(\boxed{16:25}\)

Q4. Find the Missing Value in Each Pair of Similar Solids.

(i) Pyramids

Given:

\[ l_1=12\text{ cm},\quad l_2=16\text{ cm},\quad V_2=1536\text{ cm}^3 \]

\[ \frac{V_1}{V_2}=\left(\frac{l_1}{l_2}\right)^3 \]

\[ \frac{V_1}{1536}=\left(\frac{12}{16}\right)^3=\frac{27}{64} \]

\[ V_1=1536\times\frac{27}{64} \]

\[ \boxed{V_1=648\text{ cm}^3} \]

(ii) Cylinders

Given:

\[ h_1=2.5\text{ m},\quad h_2=8.75\text{ m},\quad V_2=171.5\text{ cm}^3 \]

The height ratio is

\[ \frac{h_1}{h_2}=\frac{2.5}{8.75}=\frac27 \]

Using the volume ratio,

\[ \frac{V_1}{171.5}=\left(\frac27\right)^3=\frac8{343} \]

\[ V_1=171.5\times\frac8{343} \]

\[ \boxed{V_1=4\text{ cm}^3} \]

Note: The given heights are in metres, while the volume is in cubic centimetres. Check the original question to confirm the units.

(iii) Cuboids

Given:

\[ A_1=392\text{ cm}^2,\quad A_2=162\text{ cm}^2,\quad V_2=729\text{ cm}^3 \]

Using the supplied surface-area ratio,

\[ \left(\frac{l_1}{l_2}\right)^2=\frac{A_1}{A_2} =\frac{392}{162}=\frac{196}{81} \]

\[ \frac{l_1}{l_2}=\sqrt{\frac{196}{81}}=\frac{14}{9} \]

Now use the volume ratio:

\[ \frac{V_1}{729}=\left(\frac{14}{9}\right)^3 =\frac{2744}{729} \]

\[ \boxed{V_1=2744\text{ cm}^3} \]

Note: This calculation assumes the given areas are comparable corresponding surface areas of similar solids.

(iv) Spheres

Given:

\[ V_1=64\text{ cm}^3,\quad V_2=216\text{ cm}^3,\quad r_2=12\text{ cm} \]

\[ \left(\frac{r_1}{12}\right)^3=\frac{64}{216} \]

\[ \frac{r_1}{12}=\sqrt[3]{\frac{64}{216}}=\frac23 \]

\[ r_1=12\times\frac23 \]

\[ \boxed{r_1=8\text{ cm}} \]

Q5. The Ratio of Corresponding Lengths of Two Similar Cans Is \(3:2\).

(i) The Larger Can Has a Surface Area of 96 m². Find the Surface Area of the Smaller Can.

\[ \frac{A_{\text{small}}}{A_{\text{large}}} =\left(\frac{l_{\text{small}}}{l_{\text{large}}}\right)^2 \]

\[ \frac{A_{\text{small}}}{96}=\left(\frac23\right)^2=\frac49 \]

\[ A_{\text{small}}=96\times\frac49 \]

\[ \boxed{A_{\text{small}}\approx42.67\text{ m}^2} \]

(ii) The Smaller Can Has a Volume of 240 m³. Find the Volume of the Larger Can.

\[ \frac{V_{\text{large}}}{V_{\text{small}}} =\left(\frac{l_{\text{large}}}{l_{\text{small}}}\right)^3 \]

\[ \frac{V_{\text{large}}}{240}=\left(\frac32\right)^3=\frac{27}{8} \]

\[ V_{\text{large}}=240\times\frac{27}{8} \]

\[ \boxed{V_{\text{large}}=810\text{ m}^3} \]

Q6. The Ratio of the Heights of Two Similar Cylindrical Water Tanks Is \(5:3\).

(i) The Larger Tank Has a Surface Area of 250 m². Find the Surface Area of the Smaller Tank.

\[ \frac{A_{\text{small}}}{A_{\text{large}}} =\left(\frac{h_{\text{small}}}{h_{\text{large}}}\right)^2 \]

\[ \frac{A_{\text{small}}}{250}=\left(\frac35\right)^2=\frac9{25} \]

\[ A_{\text{small}}=250\times\frac9{25} \]

\[ \boxed{A_{\text{small}}=90\text{ m}^2} \]

(ii) The Smaller Tank Has a Volume of 270 m³. Find the Volume of the Larger Tank.

\[ \frac{V_{\text{large}}}{V_{\text{small}}} =\left(\frac{h_{\text{large}}}{h_{\text{small}}}\right)^3 \]

\[ \frac{V_{\text{large}}}{270}=\left(\frac53\right)^3=\frac{125}{27} \]

\[ V_{\text{large}}=270\times\frac{125}{27} \]

\[ \boxed{V_{\text{large}}=1250\text{ m}^3} \]

Exercise 9.4 Solutions

Q1. Polygon Angle Calculations

(i) Find the Sum of the Interior Angles of a Decagon.

For an \(n\)-sided polygon,

\[ S=(n-2)\times180^\circ \]

For a decagon, \(n=10\).

\[ S=(10-2)\times180^\circ \]

\[ =8\times180^\circ \]

\[ \boxed{1440^\circ} \]

(ii) Calculate Each Interior Angle of a Regular Hexagon.

\[ \text{Each interior angle}=\frac{(n-2)\times180^\circ}{n} \]

For \(n=6\),

\[ \frac{(6-2)\times180^\circ}{6} =\frac{720^\circ}{6} \]

\[ \boxed{120^\circ} \]

(iii) Find Each Exterior Angle of a Regular Pentagon.

\[ \text{Exterior angle}=\frac{360^\circ}{n} \]

\[ =\frac{360^\circ}{5} \]

\[ \boxed{72^\circ} \]

(iv) A Polygon Has an Interior Angle Sum of \(1440^\circ\). Find the Number of Sides.

\[ (n-2)\times180^\circ=1440^\circ \]

\[ n-2=\frac{1440}{180}=8 \]

\[ n=8+2 \]

\[ \boxed{10\text{ sides}} \]

Q2. In a Parallelogram \(ABCD\), \(AB=10\) cm, \(AD=6\) cm, and \(\angle BAD=45^\circ\). Calculate the Length of Diagonal \(AC\).

In a parallelogram, opposite sides are equal.

\[ BC=AD=6\text{ cm} \]

The adjacent angle is

\[ \angle ABC=180^\circ-45^\circ=135^\circ \]

Using the cosine rule in \(\triangle ABC\),

\[ AC^2=AB^2+BC^2-2(AB)(BC)\cos135^\circ \]

\[ AC^2=10^2+6^2-2(10)(6)\cos135^\circ \]

Since \(\cos135^\circ\approx-0.7071\),

\[ AC^2\approx100+36+84.85 \]

\[ AC^2\approx220.85 \]

\[ AC\approx\sqrt{220.85} \]

\[ \boxed{AC\approx14.86\text{ cm}} \]

Q3. In a Parallelogram \(ABCD\), If \(\angle DAB=70^\circ\), Find All the Other Angles.

Opposite angles of a parallelogram are equal.

\[ \angle BCD=\angle DAB=70^\circ \]

Adjacent angles are supplementary.

\[ \angle ABC=180^\circ-70^\circ=110^\circ \]

The opposite angle is

\[ \angle ADC=\angle ABC=110^\circ \]

Answers:

\[ \boxed{\angle BCD=70^\circ} \]

\[ \boxed{\angle ABC=110^\circ} \]

\[ \boxed{\angle ADC=110^\circ} \]

Q4. A Shape Is Created by Cutting a Square in Half Diagonally and Attaching a Right-Angled Triangle to the Hypotenuse of Each Half. Explain Why It Can Tessellate and Calculate the Interior Angles.

Explanation: A shape can tessellate if copies fit together without gaps or overlaps. The angles meeting at a point must add up to \(360^\circ\).

Cutting a square diagonally forms two isosceles right triangles with angles

\[ 45^\circ,\quad45^\circ,\quad90^\circ \]

The supplied solution describes the resulting shape as having interior angles

\[ \boxed{90^\circ,\ 90^\circ,\ 135^\circ,\ 135^\circ} \]

Note: The exact shape and its angles depend on the dimensions and orientation of the attached triangles. Check the original diagram to confirm this result.

Q5. A Tessellation Uses a Right-Angled Triangle with Sides 3, 4, and 5 Units. Find the Minimum Number of Reflections Needed to Cover a Square of Area 3600 Square Units.

Step 1: Find the area of the triangle.

The perpendicular sides are 3 and 4 units.

\[ A=\frac12\times3\times4=6\text{ square units} \]

Step 2: Find the number of triangles required.

\[ N=\frac{3600}{6}=600 \]

Step 3: Calculate the number of reflections.

Using the supplied assumption that each reflection adds one triangle,

\[ 600-1=\boxed{599\text{ reflections}} \]

Q6. A Tessellation Uses 25 Regular Hexagons, Each with Side Length 5 cm. Find the Total Area and the Outer Perimeter, Assuming the Outer Shape Is a Perfect Hexagon.

Step 1: Find the area of one hexagon.

\[ A=\frac{3\sqrt3}{2}s^2 \]

\[ A=\frac{3\sqrt3}{2}(5)^2 \]

\[ A=\frac{75\sqrt3}{2}\approx64.95\text{ cm}^2 \]

Step 2: Find the total area.

\[ A_{\text{total}}=25\times64.95 \]

\[ \boxed{A_{\text{total}}\approx1623.8\text{ cm}^2} \]

Step 3: Calculate the outer perimeter using the supplied assumption.

If the outer regular hexagon has side length 25 cm,

\[ P=6\times25 \]

\[ \boxed{P=150\text{ cm}} \]

Note: Twenty-five small hexagons do not automatically form a single regular hexagon with a linear scale factor of 5. The outer perimeter depends on the actual arrangement, so check the original diagram.

Q7. A Rectangular Floor Is 12 m by 15 m. How Many Square Tiles of Size 1 m by 1 m Are Needed?

\[ \text{Floor area}=12\times15=180\text{ m}^2 \]

\[ \text{Area of one tile}=1\times1=1\text{ m}^2 \]

\[ \text{Number of tiles}=\frac{180}{1} \]

\[ \boxed{180\text{ tiles}} \]

Q8. A Rectangular Wall Is 10 m Tall and 12 m Wide. How Many Gallons of Paint Are Needed if One Gallon Covers 350 m²?

\[ \text{Wall area}=10\times12=120\text{ m}^2 \]

\[ \text{Gallons required}=\frac{120}{350} \]

\[ \boxed{\approx0.34\text{ gallons}} \]

This is the calculated amount; the quantity purchased may need to be rounded up to a whole gallon.

Q9. A Rectangular Wall Is 10 m Long and 4 m Wide. If One Litre of Paint Covers 2 m², How Many Litres Are Needed?

\[ \text{Wall area}=10\times4=40\text{ m}^2 \]

\[ \text{Paint required}=\frac{40}{2} \]

\[ \boxed{20\text{ litres}} \]

Q10. A Trapezoidal Window Has Parallel Sides of 3 m and 1.5 m and a Height of 2 m. Find Its Area.

The area of a trapezoid is

\[ A=\frac12(a+b)h \]

Substitute the given values:

\[ A=\frac12(3+1.5)\times2 \]

\[ A=\frac12(4.5)\times2 \]

\[ \boxed{4.5\text{ m}^2} \]

Review Exercise 9 Solutions

Q1. Multiple Choice Questions

(i) If Two Polygons Are Similar, Then:

Correct Answer: (a) Their corresponding angles are equal.

Explanation: Similar polygons have equal corresponding angles and proportional corresponding sides.

(ii) The Ratio of the Areas of Two Similar Polygons Is:

Correct Answer: (b) Equal to the square of the ratio of their corresponding sides.

\[ \frac{A_1}{A_2}=\left(\frac{l_1}{l_2}\right)^2 \]

(iii) If the Volumes of Two Similar Solids Are 125 cm³ and 27 cm³, Find the Ratio of Their Corresponding Heights.

\[ \frac{V_1}{V_2}=\left(\frac{h_1}{h_2}\right)^3 \]

\[ \frac{h_1}{h_2}=\sqrt[3]{\frac{125}{27}}=\frac53 \]

Correct Answer: (b) \(5:3\)

(iv) The Exterior Angle of a Regular Pentagon Is:

\[ \frac{360^\circ}{5}=72^\circ \]

Correct Answer: (d) \(72^\circ\)

(v) A Smaller Parallelogram Has an Area of 64 cm² and a Similar Larger Parallelogram Has an Area of 144 cm². If a Side of the Smaller Parallelogram Is 8 cm, Find the Corresponding Side of the Larger Parallelogram.

\[ \left(\frac{s_2}{s_1}\right)^2=\frac{A_2}{A_1} \]

\[ \left(\frac{s_2}{8}\right)^2=\frac{144}{64} \]

\[ \frac{s_2}{8}=\sqrt{\frac{144}{64}}=\frac{12}{8} \]

\[ s_2=12\text{ cm} \]

Correct Answer: (b) 12 cm

(vi) Find the Total Number of Diagonals in a Polygon with 9 Sides.

The formula is

\[ D=\frac{n(n-3)}2 \]

\[ D=\frac{9(9-3)}2 \]

\[ =\frac{9\times6}{2} \]

\[ D=27 \]

Correct Answer: (d) 27

(vii) Two Spheres Have Radii in the Ratio \(4:5\). If the Surface Area of the Larger Sphere Is \(500\pi\) cm², Find the Surface Area of the Smaller Sphere.

\[ \frac{A_1}{A_2}=\left(\frac{r_1}{r_2}\right)^2 \]

\[ \frac{A_1}{500\pi}=\left(\frac45\right)^2=\frac{16}{25} \]

\[ A_1=500\pi\times\frac{16}{25} \]

\[ \boxed{A_1=320\pi\text{ cm}^2} \]

Correct Answer: (b) \(320\pi\text{ cm}^2\)

(viii) A Regular Polygon Has an Exterior Angle of \(30^\circ\). Find the Number of Diagonals.

\[ n=\frac{360^\circ}{30^\circ}=12 \]

\[ D=\frac{n(n-3)}2 \]

\[ D=\frac{12(12-3)}2 \]

\[ =\frac{12\times9}{2} \]

\[ D=54 \]

Correct Answer: (a) 54

(ix) In a Regular Hexagon, Find the Ratio of the Longest Diagonal to the Side Length.

The longest diagonal of a regular hexagon is twice its side length.

\[ \frac{d}{s}=\frac{2s}{s}=2 \]

Correct Answer: (b) \(2:1\)

(x) A Regular Polygon Has an Interior Angle of \(165^\circ\). Find the Number of Sides.

\[ \text{Exterior angle}=180^\circ-165^\circ=15^\circ \]

\[ n=\frac{360^\circ}{15^\circ}=24 \]

Correct Answer: (d) 24

Q2. If the Sum of the Interior Angles of a Polygon Is \(1080^\circ\), Find the Number of Sides.

\[ (n-2)\times180^\circ=1080^\circ \]

\[ n-2=\frac{1080}{180}=6 \]

\[ n=6+2 \]

\[ \boxed{8\text{ sides}} \]

The polygon is an octagon.

Q3. Two Similar Bottles Have Heights in the Ratio \(2:1\). Find the Ratios of Their Surface Areas and Capacities.

(a) Ratio of Surface Areas

\[ \frac{A_{\text{large}}}{A_{\text{small}}} =\left(\frac21\right)^2 \]

\[ \boxed{4:1} \]

(b) Ratio of Capacities

\[ \frac{V_{\text{large}}}{V_{\text{small}}} =\left(\frac21\right)^3 \]

\[ \boxed{8:1} \]

Q4. Each Dimension of a Model Car Is \(\frac1{10}\) of the Corresponding Dimension of the Actual Car. Find the Ratios.

(a) Areas of Their Windscreens

\[ \left(\frac1{10}\right)^2=\boxed{1:100} \]

(b) Capacities of Their Boots

\[ \left(\frac1{10}\right)^3=\boxed{1:1000} \]

(c) Widths of the Cars

\[ \boxed{1:10} \]

(d) Number of Wheels

Both cars have the same number of wheels.

\[ \boxed{1:1} \]

Q5. Three Similar Jugs Have Heights of 8 cm, 12 cm, and 16 cm. If the Smallest Jug Holds 0.5 L, Find the Capacities of the Other Two.

(i) Capacity of the 12 cm Jug

\[ \frac{V_2}{V_1}=\left(\frac{h_2}{h_1}\right)^3 \]

\[ \frac{V_2}{0.5}=\left(\frac{12}{8}\right)^3 =\left(\frac32\right)^3=\frac{27}{8} \]

\[ V_2=0.5\times\frac{27}{8} \]

\[ \boxed{V_2=1.6875\text{ L}} \]

(ii) Capacity of the 16 cm Jug

\[ \frac{V_3}{0.5}=\left(\frac{16}{8}\right)^3 =2^3=8 \]

\[ V_3=0.5\times8 \]

\[ \boxed{V_3=4\text{ L}} \]

Q6. Three Similar Drinking Glasses Have Heights of 7.5 cm, 9 cm, and 10.5 cm. If the Tallest Glass Holds 343 mL, Find the Capacities of the Other Two.

(i) Capacity of the 7.5 cm Glass

\[ \frac{V_1}{343}=\left(\frac{7.5}{10.5}\right)^3 =\left(\frac57\right)^3=\frac{125}{343} \]

\[ V_1=343\times\frac{125}{343} \]

\[ \boxed{V_1=125\text{ mL}} \]

(ii) Capacity of the 9 cm Glass

\[ \frac{V_2}{343}=\left(\frac9{10.5}\right)^3 =\left(\frac67\right)^3=\frac{216}{343} \]

\[ V_2=343\times\frac{216}{343} \]

\[ \boxed{V_2=216\text{ mL}} \]

Q7. A Model Car Is Similar to the Actual Car. The Ratio of the Model’s Door Area to the Actual Car’s Door Area Is \(1:2500\). Find the Following.

(a) Ratio of Their Lengths

\[ \frac{l_{\text{model}}}{l_{\text{car}}} =\sqrt{\frac{A_{\text{model}}}{A_{\text{car}}}} \]

\[ =\sqrt{\frac1{2500}}=\frac1{50} \]

Answer: \(\boxed{1:50}\)

(b) Ratio of the Capacities of Their Petrol Tanks

\[ \frac{V_{\text{model}}}{V_{\text{car}}} =\left(\frac1{50}\right)^3 \]

\[ \boxed{1:125000} \]

(c) Width of the Model if the Actual Car Is 150 cm Wide

\[ w_{\text{model}}=\frac1{50}\times150 \]

\[ \boxed{3\text{ cm}} \]

(d) Area of the Actual Car’s Rear Window if the Model’s Rear Window Has an Area of 3 cm²

\[ A_{\text{car}}=2500\times A_{\text{model}} \]

\[ =2500\times3 \]

\[ \boxed{7500\text{ cm}^2} \]

Q8. The Ratio of the Areas of Two Similar Labels on Two Similar Coffee Jars Is \(144:169\). Find the Ratios of Their Heights and Capacities.

(a) Ratio of Heights

\[ \frac{h_1}{h_2}=\sqrt{\frac{A_1}{A_2}} \]

\[ =\sqrt{\frac{144}{169}}=\frac{12}{13} \]

Answer: \(\boxed{12:13}\)

(b) Ratio of Capacities

\[ \frac{V_1}{V_2}=\left(\frac{12}{13}\right)^3 \]

\[ =\frac{1728}{2197} \]

Answer: \(\boxed{1728:2197}\)

Q9. A Tessellation Uses One Regular Hexagon, Six Squares, and Six Equilateral Triangles. Each Polygon Has Side Length \(\frac12\) m. Find the Total Area of One Pattern.

Given:

\[ s=\frac12\text{ m}=0.5\text{ m} \]

Step 1: Find the Area of One Regular Hexagon.

\[ A_{\text{hexagon}}=\frac{3\sqrt3}{2}s^2 \]

\[ =\frac{3\sqrt3}{2}\left(\frac12\right)^2 \]

\[ =\frac{3\sqrt3}{8}\text{ m}^2 \]

Step 2: Find the Total Area of Six Squares.

Area of one square:

\[ A_{\text{square}}=s^2=\left(\frac12\right)^2=\frac14\text{ m}^2 \]

Area of six squares:

\[ 6\times\frac14=\frac32\text{ m}^2 \]

Step 3: Find the Total Area of Six Equilateral Triangles.

Area of one equilateral triangle:

\[ A_{\text{triangle}}=\frac{\sqrt3}{4}s^2 \]

\[ =\frac{\sqrt3}{4}\left(\frac12\right)^2 =\frac{\sqrt3}{16}\text{ m}^2 \]

Area of six triangles:

\[ 6\times\frac{\sqrt3}{16} =\frac{3\sqrt3}{8}\text{ m}^2 \]

Step 4: Find the Total Area.

\[ A_{\text{total}}=A_{\text{hexagon}}+A_{\text{squares}}+A_{\text{triangles}} \]

\[ =\frac{3\sqrt3}{8}+\frac32+\frac{3\sqrt3}{8} \]

\[ =\frac{3\sqrt3}{4}+\frac32 \]

Therefore,

\[ \boxed{A_{\text{total}}=\frac{3\sqrt3}{4}+\frac32\text{ m}^2} \]

Using \(\sqrt3\approx1.73205\),

\[ A_{\text{total}}\approx1.299+1.5 \]

\[ \boxed{A_{\text{total}}\approx2.80\text{ m}^2} \]

Notes by Sir Dilawar

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